How do you find the inflection points of the graph of the function: #f(x)= (x^7/42) - (3x^6/10) + (6x^5/5) - (4x^4/3)#?

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Answer 1 · 2015-10-18T00:49:22

See the explanation.

For #f(x)= (x^7/42) - (3x^6/10) + (6x^5/5) - (4x^4/3)#, we get

#f''(x) = x^5-9x^4+24x^3-16x^2#

# = x^2( x^3-9x^2+24x-16)#

Looking for rational zeros of the cubic, we note that #1# is a zro, so #x-1# is a factor. Dividing by #x-1# gets us:

# = x^2(x-1) (x^2-8x+16)#

# = x^2(x-1) (x-4)^2)#

The zeros of #f''# are #0#, #1#, and #4#.

The only one at which #f''# changes sign is #1#.

The only inflection point is #(1,f(1))#

(I'll leave it to the student to find #f(1)#)